Đáp án:
$\begin{array}{l}
1){x^2} = \left( {2m - 1} \right)x + 2m\\
\Leftrightarrow {x^2} - \left( {2m - 1} \right).x - 2m = 0\\
\Leftrightarrow \Delta > 0\\
\Leftrightarrow {\left( {2m - 1} \right)^2} - 4.\left( { - 2m} \right) > 0\\
\Leftrightarrow 4{m^2} - 4m + 1 + 8m > 0\\
\Leftrightarrow 4{m^2} + 4m + 1 > 0\\
\Leftrightarrow {\left( {2m + 1} \right)^2} > 0\\
\Leftrightarrow m\# - \dfrac{1}{2}\\
Vậy\,m\# - \dfrac{1}{2}\\
2)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 1\\
{x_1}{x_2} = - 2m
\end{array} \right.\\
Khi:\left| {{x_1}} \right| = \left| {{x_2}} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
{x_1} = {x_2}\left( {ktm} \right)\\
{x_1} = - {x_2}\left( {tm} \right)
\end{array} \right.\\
\Leftrightarrow {x_1} + {x_2} = 0\\
\Leftrightarrow 2m - 1 = 0\\
\Leftrightarrow m = \dfrac{1}{2}\left( {tm} \right)\\
Vậy\,m = \dfrac{1}{2}
\end{array}$
